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In an irreversible process taking place at constant \(\mathrm{T}\) and \(\mathrm{P}\) and in which only pressure -volume work is being done, the change in Gibbs free energy \((\mathrm{dG})\) and change in entropy (dS), satisfy the criteria: a. \((\mathrm{dS})_{\mathrm{V}, \mathrm{E}}<0,(\mathrm{dG})_{\mathrm{T}, \mathrm{P}}<0\) b. \((\mathrm{dS})_{\mathrm{V}, \mathrm{E}}>0,(\mathrm{~d} \mathrm{G})_{\mathrm{T}, \mathrm{P}}<0\) c. \((\mathrm{dS})_{\mathrm{V}, \mathrm{E}}=0,(\mathrm{dG})_{\mathrm{T}, \mathrm{P}}=0\) d. \((\mathrm{dS})_{\mathrm{V}, \mathrm{E}}=0,(\mathrm{dG})_{\mathrm{T}, \mathrm{P}}>0\)

Short Answer

Expert verified
The correct choice is option (b): \((\mathrm{dS}) > 0, (\mathrm{dG}) < 0\).

Step by step solution

01

Define the conditions of the irreversible process

In this problem, we have an irreversible process occurring at constant temperature (T) and pressure (P), with only pressure-volume work being done. We need to evaluate how the changes in Gibbs free energy \((\mathrm{dG})\) and entropy \((\mathrm{dS})\) behave under these conditions.
02

Apply the properties of Gibbs free energy

For a spontaneous process at constant temperature and pressure, the Gibbs free energy \((\mathrm{dG})\) should decrease, which means \((\mathrm{dG}) < 0\). This is a fundamental thermodynamic criterion for spontaneity in isothermal and isobaric conditions.
03

Apply the properties of entropy change in irreversible processes

In an irreversible process, the entropy of the system plus its surroundings must increase, so \((\mathrm{dS}) > 0\). This is in accordance with the second law of thermodynamics.
04

Determine the correct option

Based on the criteria for Gibbs free energy and entropy: - Gibbs free energy change is negative, \((\mathrm{dG}) < 0\), and- Entropy change is positive, \((\mathrm{dS}) > 0\). The correct choice is option (b): \((\mathrm{dS})_{\mathrm{V}, \mathrm{E}}>0, (\mathrm{dG})_{\mathrm{T}, \mathrm{P}}<0\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gibbs Free Energy
Gibbs free energy, often denoted as \( G \), is an essential concept in thermodynamics, particularly when discussing spontaneous processes. It combines the concepts of enthalpy, entropy, and temperature. The formula for Gibbs free energy is given by \( G = H - TS \), where \( H \) is enthalpy, \( T \) is temperature, and \( S \) is entropy.
Gibbs free energy is a measure of the maximum reversible work that a system can perform at constant temperature and pressure. When we talk about changes in \( G \), we usually refer to whether a process can occur spontaneously. The change in Gibbs free energy, \( \mathrm{dG} \), helps us predict the direction of a process.
  • If \( \mathrm{dG} < 0 \), the process is spontaneous.
  • If \( \mathrm{dG} = 0 \), the system is in equilibrium.
  • If \( \mathrm{dG} > 0 \), the process is non-spontaneous.
In the context of the exercise, for an irreversible process at constant \( T \) and \( P \), \( \mathrm{dG} \) must be negative for the process to be spontaneous, supporting option b.
Entropy Change
Entropy, usually symbolized as \( S \), is a measure of disorder or randomness in a system. It reflects the number of ways the system can be arranged at a microscopic level.
According to the second law of thermodynamics, the total entropy of an isolated system can never decrease; it either increases or remains constant. This principle helps us understand irreversible processes. In an irreversible process, the entropy change of the universe \( (dS_{universe}) \) is always greater than zero.
Within the exercise, we recognize that the irreversible process occurring causes a net increase in entropy, meaning \( \mathrm{dS} > 0 \). This aligns perfectly with the second law and reinforces why option b is the correct choice for the problem. An increase in entropy reflects the natural tendency towards disorder, a key aspect of irreversible processes.
Thermodynamics
Thermodynamics is the study of energy, heat, and work in physical systems. The laws of thermodynamics provide a statistical account of how and why energy transfer occurs, and what limitations exist for the processes.
There are four laws of thermodynamics which govern the energetics of all physical systems:
  • **Zeroth Law**: If two systems are in thermal equilibrium with a third system, they are in thermal equilibrium with each other.
  • **First Law**: Energy cannot be created or destroyed, only transferred or converted. This is often summarized as the principle of the conservation of energy.
  • **Second Law**: The total entropy of an isolated system can never decrease over time. It highlights the directional flow of energy transformations.
  • **Third Law**: As the temperature of a system approaches absolute zero, the entropy of the system approaches a constant minimum.
In reference to the exercise, understanding the second law is crucial, as it dictates that entropy increases in irreversible processes, providing the solution needed for choosing the correct option.

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Most popular questions from this chapter

One mole of a non-ideal gas undergoes a change of state \((2.0 \mathrm{~atm}, 3.0 \mathrm{~L}, 95 \mathrm{~K}) \rightarrow(4.0 \mathrm{~atm}, 5.0 \mathrm{~L}, 245 \mathrm{~K})\) with a change in internal energy, \(\Delta \mathrm{U}=30.0 \mathrm{~L}\) atm. The change in enthalpy \((\Delta \mathrm{H})\) of the process in \(\mathrm{L}\) atm is a. \(40.0\) b. \(42.3\) c. \(44.0\) d. not defined, because pressure is not constant [IIT 2002]

For which of the following processes will the entropy decrease? a. Reaction of magnesium with oxygen to form magnesium oxide. b. Reaction of nitrogen and hydrogen to form ammonia. c. Sublimation of dry ice. d. Condensation of steam.

(A): Chlorine when tried to be solidified does not have zero entropy when at absolute zero. (R): Chlorine contains a mixture of isotopes and is difficult to solidify.

Identify the incorrect statement/s regarding entropy. a. at absolute zero temperature, entropy of perfectly crystalline substance is taken to zero b. at absolute zero temperature, the entropy of perfectly crystalline substance is positive c. at \(0^{\circ} \mathrm{C}\) the entropy of perfectly crystalline substance is positive d. at absolute zero temperature, the entropy of all crystalline substances is taken to be zero

Which of the following reaction defines \(\Delta \mathrm{H}^{\circ}\) ? a. \(\mathrm{C}\) (diamond) \(+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})\) b. \(1 / 2 \mathrm{H}_{2}(\mathrm{~g})+1 / 2 \mathrm{~F}_{2}(\mathrm{~g}) \rightarrow \mathrm{HF}(\mathrm{g})\) c. \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\) d. \(\mathrm{CO}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})\) [IIT 2003]

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